16.5 Surface Integrals of Vector Fields

Transcription

1 16.5 Surface Integrals of Vector Fields Lukas Geyer Montana State University M73, Fall 011 Lukas Geyer (MSU) 16.5 Surface Integrals of Vector Fields M73, Fall / 19

2 Parametrized Surfaces Definition An orientation on a surface S is a continuous choice of a unit normal vector e n (P) at each point P os S. The xy-plane has two orientations, one given by e n = k (pointing up), the other by e n = k (pointing down). The sphere x = R has two orientations, one given by the outward pointing vector e n (x) = x x, the other by the inward pointing normal vectors e n (x). Lukas Geyer (MSU) 16.5 Surface Integrals of Vector Fields M73, Fall 011 / 19

3 The Möbius Strip Caution Not all surfaces are orientable. The most popular example of a non-orientable surface is the Möbius strip depicted below. Lukas Geyer (MSU) 16.5 Surface Integrals of Vector Fields M73, Fall / 19

4 Vector Surface Integral Definition The normal component of a vector field F at a point P on an oriented surface S is F(P) e n (P) Definition The vector surface integral of a vector field F over a surface S is F ds = (F e n )ds. It is also called the flux of F across or through S. Applications S Flow rate of a fluid with velocity field F across a surface S. Magnetic and electric flux across surfaces. (Maxwell s equations) S Lukas Geyer (MSU) 16.5 Surface Integrals of Vector Fields M73, Fall / 19

5 Parametrized Vector Surface Integral Calculating Parametrized Surface Integrals A regular parametrization G(u, v) (i.e., a parametrization with n(u, v) 0 for all u, v) induces an orientation by e n =. We get S Differential Version F S = D n n F(G(u, v)) n(u, v) du dv. ds = n(u, v) du dv Orientation matters Reversing the orientation of S changes the sign of the integral. Lukas Geyer (MSU) 16.5 Surface Integrals of Vector Fields M73, Fall / 19

6 Vector Surface Integral I Find F ds, where F = 0, 0, z and S is the oriented surface S parametrized by G(u, v) = (u cos v, u sin v, v), where 0 u 1, 0 v π. T u = cos v, sin v, 0 T v = u sin v, u cos v, 1 n = T u T v = sin v, cos v, u Is the integral positive, negative, or 0? Positive! Lukas Geyer (MSU) 16.5 Surface Integrals of Vector Fields M73, Fall / 19

7 Vector Surface Integral II Find F ds, where F = 0, 0, z and S is the oriented surface S parametrized by G(u, v) = (u cos v, u sin v, v), where 0 u 1, 0 v π. Evaluating the integral S n = sin v, cos v, u F ds = = 1 π π 0 0 = 1 4π = π. 0, 0, v sin v, cos v, u dv du ( 1 vu dv du = 0 ) ( π ) u du v dv 0 Lukas Geyer (MSU) 16.5 Surface Integrals of Vector Fields M73, Fall / 19

8 Flow Rate I A fluid flows with constant velocity v = 3k(m/s). Calculate the flow rate (in m 3 /s) through the part of the elliptic paraboloid z = x + y with z 4 and upward pointing normal vector. Parametrize Surface At height z the trace is a circle of radius r = z. Using r and θ as parameters: G(r, θ) = (r cos θ, r sin θ, r ), 0 r, 0 θ π T r = cos θ, sin θ, r T θ = r sin θ, r cos θ, 0 n = T r T θ = r cos θ, r sin θ, r Lukas Geyer (MSU) 16.5 Surface Integrals of Vector Fields M73, Fall / 19

9 Flow Rate II A fluid flows with constant velocity v = 3k(m/s). Calculate the flow rate (in m 3 /s) through the part of the elliptic paraboloid z = x + y with z 4 and upward pointing normal vector. Integrate G(r, θ) = (r cos θ, r sin θ, r ), 0 r, 0 θ π n = r cos θ, r sin θ, r S F ds = = π 0 0 π 0 0 0, 0, 3 r cos θ, r sin θ, r dθ dr 3r dθ dr = 6π 0 r dr = 1π. Lukas Geyer (MSU) 16.5 Surface Integrals of Vector Fields M73, Fall / 19

10 Another I Let S be the boundary of the solid cone with base x + y 4 in the Parametrizing S We have to parametrize both the base and the boundary cone, calculate the flux through both and add the results. Parametrizing the base Standard parametrization of a disk of radius : G(r, θ) = (r cos θ, r sin θ, 0), 0 r, 0 θ π. Lukas Geyer (MSU) 16.5 Surface Integrals of Vector Fields M73, Fall / 19

11 Another II Let S be the boundary of the solid cone with base x + y 4 in the Parametrizing the base G(r, θ) = (r cos θ, r sin θ, 0), 0 r, 0 θ π, T r = cos θ, sin θ, 0 T θ = r sin θ, r cos θ, 0 n = T r T θ = 0, 0, r Problem: r 0, so n points up, into the cone, not out. Solution: Change the sign of n. Lukas Geyer (MSU) 16.5 Surface Integrals of Vector Fields M73, Fall / 19

12 Another III Let S be the boundary of the solid cone with base x + y 4 in the Integrating over the base G(r, θ) = (r cos θ, r sin θ, 0), 0 r, 0 θ π, S 1 F ds = n = 0, 0, r π 0 0 r sin θ, r cos θ, 0 0, 0, r dθ dr = 0 Lukas Geyer (MSU) 16.5 Surface Integrals of Vector Fields M73, Fall / 19

13 Another IV Let S be the boundary of the solid cone with base x + y 4 in the Parametrizing the cone Trace in z = k is a circle of radius (4 k)/ for 0 k 4, so 4 z G(z, θ) = cos θ, 4 z sin θ, z, 0 z 4, 0 θ π. T z = 1 cos θ, 1 sin θ, 1 T θ = 4 z sin θ, 4 z cos θ, 0 z 4 n = T z T θ = cos θ, z 4 sin θ, z 4. Lukas Geyer (MSU) 16.5 Surface Integrals of Vector Fields M73, Fall / 19

14 Another V Let S be the boundary of the solid cone with base x + y 4 in the Integrating over the cone G(z, θ) = n = 4 z z 4 cos θ, 4 z cos θ, z 4 sin θ, z, 0 z 4, 0 θ π. sin θ, z 4. (z 4)/ 0, so n points down, i.e., into the cone. Again we have to change the sign to fix the orientation. Lukas Geyer (MSU) 16.5 Surface Integrals of Vector Fields M73, Fall / 19

15 Another VI Let S be the boundary of the solid cone with base x + y 4 in the Integrating over the cone, correct orientation G(z, θ) = n = 4 z 4 z cos θ, 4 z cos θ, 4 z sin θ, z, 0 z 4, 0 θ π. sin θ, 4 z. z 4 F n = sin θ, 4 z 4 z cos θ, z cos θ, 4 z sin θ, 4 z Lukas Geyer (MSU) 16.5 Surface Integrals of Vector Fields M73, Fall / 19

16 Another VII Let S be the boundary of the solid cone with base x + y 4 in the Integrating over the cone, correct orientation G(z, θ) = 4 z cos θ, 4 z sin θ, z, 0 z 4, 0 θ π. z 4 F n = = z(4 z) sin θ, 4 z = 4z z 4 z cos θ, z cos θ, 4 z sin θ, 4 z Lukas Geyer (MSU) 16.5 Surface Integrals of Vector Fields M73, Fall / 19

17 Another VIII Let S be the boundary of the solid cone with base x + y 4 in the Integrating over the cone, correct orientation G(z, θ) = F n = 4 z 4z z cos θ, 4 z sin θ, z, 0 z 4, 0 θ π. S F ds = 4 π 0 0 F n dθ dz = 4 π 0 0 4z z dθ dz Lukas Geyer (MSU) 16.5 Surface Integrals of Vector Fields M73, Fall / 19

18 Another IX Let S be the boundary of the solid cone with base x + y 4 in the Integrating over the cone, correct orientation S F ds = = π 4 π 0 0 [z z3 3 4z z dθ dz = π ] z=4 z=0 4 0 ( = π z z dz ) = 3π 3. Lukas Geyer (MSU) 16.5 Surface Integrals of Vector Fields M73, Fall / 19

19 Another X Let S be the boundary of the solid cone with base x + y 4 in the Adding up the results S F ds = F ds + S 1 F ds S = 0 + 3π 3 = 3π 3. Lukas Geyer (MSU) 16.5 Surface Integrals of Vector Fields M73, Fall / 19